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3 years ago

11

AB is rotated 120° clockwise about B. Then AB transformations? is rotated 45° counterclockwise about A. What is the image of A a

s a composition of O A. (120°, B) [(-45. A))(A) B. (-45°, A ' ((120°, 3))(A) C. (r-120°. B) (45. A))(A) D. (r(45°, A) (-120°. 3))(A)

1 answer:

RideAnS [48]3 years ago

3 0

The answer to ur question is A

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Write m=-3 and (3,-1) in pointe slope form

Lady_Fox [76]

(
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3 years ago

I need help with this question. I keep messing up on the whole problem.

mylen [45]

Answer:

B

Step-by-step explanation:

Since C is included in the answers to represent that the line and be moved anywhere, you only need to find the slope.

divide the change in y by the change in x

(6-0) / (0+2) = 6/2 = 3

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3 years ago

Given: ABCD trapezoid, BK ⊥ AD , AB=DC AB=8, AK=4 Find: m∠A, m∠B

Elis [28]

Answer:

$m\angle B=m\angle C=120^{\circ}$

$m\angle A=m\angle D=60^{\circ}$

Step-by-step explanation:

Trapezoid ABCD is isosceles trapezoid, because AB = CD (given). In isosceles trapezoid, angles adjacent to the bases are congruent, then

$\angle A\cong \angle D;$
$\angle B\cong \angle C.$

Since BK ⊥ AD, the triangle ABK is right triangle. In this triangle, AB = 8, AK = 4. Note that the hypotenuse AB is twice the leg AK:

$AB=2AK.$

If in the right triangle the hypotenuse is twice the leg, then the angle opposite to this leg is 30°, so,

$m\angle ABK=30^{\circ}$

Since BK ⊥ AD, then BK ⊥ BC and

$m\angle KBC=90^{\circ}$

Thus,

$m\angle B=30^{\circ}+90^{\circ}=120^{\circ}\\ \\m\angle B=m\angle C=120^{\circ}$

Now,

$m\angle A=m\angle D=180^{\circ}-120^{\circ}=60^{\circ}$

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3 years ago

9. The difference between the length and width of a rectangle is 7 units. The perimeter is 50 units. Find the dimensions.

Mazyrski [523]

Answer:

9. 9&16

10. 31&45

Step-by-step explanation:

9. L-W=7

2L+2W=50 (system of equations)

L=W+7

2L=-2W+50

L=-W+25

-W+25=W+7

-2W=-18

W=9

L=16

10. x=larger number y=smaller number

x+y=76

x=y+14

x=-y+76

y+14=-y+76

2y=62

y=31

x=45

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3 years ago